3.212 \(\int \frac {(c+a^2 c x^2)^{3/2} \tan ^{-1}(a x)}{x} \, dx\)

Optimal. Leaf size=281 \[ -\frac {7}{6} c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )+\frac {i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 c^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {1}{6} a c x \sqrt {a^2 c x^2+c}+c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+\frac {1}{3} \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) \]

[Out]

1/3*(a^2*c*x^2+c)^(3/2)*arctan(a*x)-7/6*c^(3/2)*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))-2*c^2*arctan(a*x)*arc
tanh((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+I*c^2*polylog(2,-(1+I*a*x)^(1/2)/(
1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-I*c^2*polylog(2,(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^
2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-1/6*a*c*x*(a^2*c*x^2+c)^(1/2)+c*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4950, 4946, 4958, 4954, 217, 206, 4930, 195} \[ \frac {i c^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {i c^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {7}{6} c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )-\frac {2 c^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {1}{6} a c x \sqrt {a^2 c x^2+c}+c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+\frac {1}{3} \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/x,x]

[Out]

-(a*c*x*Sqrt[c + a^2*c*x^2])/6 + c*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/3 - (
2*c^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (7*c^(3/2)
*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/6 + (I*c^2*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1
- I*a*x])])/Sqrt[c + a^2*c*x^2] - (I*c^2*Sqrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c
+ a^2*c*x^2]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)^(
m + 1)*Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/(f*(m + 2)), x] + (Dist[d/(m + 2), Int[((f*x)^m*(a + b*ArcTan[c*x]
))/Sqrt[d + e*x^2], x], x] - Dist[(b*c*d)/(f*(m + 2)), Int[(f*x)^(m + 1)/Sqrt[d + e*x^2], x], x]) /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && NeQ[m, -2]

Rule 4950

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[
d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] + Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d + e*
x^2)^(q - 1)*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[q, 0] &&
IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

Rule 4954

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(-2*(a + b*ArcTan[c
*x])*ArcTanh[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/Sqrt[d], x] + (Simp[(I*b*PolyLog[2, -(Sqrt[1 + I*c*x]/Sqrt[1 -
I*c*x])])/Sqrt[d], x] - Simp[(I*b*PolyLog[2, Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]])/Sqrt[d], x]) /; FreeQ[{a, b, c,
d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 4958

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + c^2*
x^2]/Sqrt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/(x*Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[e, c^2*d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x} \, dx &=c \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x} \, dx+\left (a^2 c\right ) \int x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=c \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {1}{3} (a c) \int \sqrt {c+a^2 c x^2} \, dx+c^2 \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx-\left (a c^2\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {1}{6} a c x \sqrt {c+a^2 c x^2}+c \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {1}{6} \left (a c^2\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx-\left (a c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )+\frac {\left (c^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {1}{6} a c x \sqrt {c+a^2 c x^2}+c \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {2 c^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )+\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {1}{6} \left (a c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )\\ &=-\frac {1}{6} a c x \sqrt {c+a^2 c x^2}+c \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {2 c^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {7}{6} c^{3/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )+\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 220, normalized size = 0.78 \[ \frac {c \sqrt {a^2 c x^2+c} \left (-a x \sqrt {a^2 x^2+1}+2 a^2 x^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+8 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+6 i \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )-6 i \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )-\sinh ^{-1}(a x)+6 \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )-6 \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )+6 \log \left (\cos \left (\frac {1}{2} \tan ^{-1}(a x)\right )-\sin \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )-6 \log \left (\sin \left (\frac {1}{2} \tan ^{-1}(a x)\right )+\cos \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )\right )}{6 \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/x,x]

[Out]

(c*Sqrt[c + a^2*c*x^2]*(-(a*x*Sqrt[1 + a^2*x^2]) - ArcSinh[a*x] + 8*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + 2*a^2*x^2*
Sqrt[1 + a^2*x^2]*ArcTan[a*x] + 6*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] - 6*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a
*x])] + 6*Log[Cos[ArcTan[a*x]/2] - Sin[ArcTan[a*x]/2]] - 6*Log[Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2]] + (6*I
)*PolyLog[2, -E^(I*ArcTan[a*x])] - (6*I)*PolyLog[2, E^(I*ArcTan[a*x])]))/(6*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \arctan \left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x)/x,x, algorithm="fricas")

[Out]

integral((a^2*c*x^2 + c)^(3/2)*arctan(a*x)/x, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x)/x,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.68, size = 174, normalized size = 0.62 \[ \frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (2 \arctan \left (a x \right ) x^{2} a^{2}-a x +8 \arctan \left (a x \right )\right )}{6}+\frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (7 i \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+3 i \dilog \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-3 \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+3 i \dilog \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{3 \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(3/2)*arctan(a*x)/x,x)

[Out]

1/6*c*(c*(a*x-I)*(I+a*x))^(1/2)*(2*arctan(a*x)*x^2*a^2-a*x+8*arctan(a*x))+1/3*c*(c*(a*x-I)*(I+a*x))^(1/2)*(7*I
*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))+3*I*dilog(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))-3*arctan(a*x)*ln(1+(1+I*a*x)/(a^
2*x^2+1)^(1/2))+3*I*dilog((1+I*a*x)/(a^2*x^2+1)^(1/2)))/(a^2*x^2+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (a^{2} c x^{2} + c\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c} \arctan \left (a x\right ) - \frac {1}{6} \, {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} {\left (a c x \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, c \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt {c} + \frac {1}{12} \, {\left (c \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) + c \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) + 12 \, c \int \frac {\sqrt {a^{2} x^{2} + 1} \arctan \left (a x\right )}{x}\,{d x}\right )} \sqrt {c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x)/x,x, algorithm="maxima")

[Out]

1/3*(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 1/6*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*c*x*cos(1/
2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) + 1/12*(c*arctan2((a^4*x^
4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos
(1/2*arctan2(4*a*x, a^2*x^2 - 3))) + c*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2
- 3)) - 2, -a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))) + 12*c*integrate(sqr
t(a^2*x^2 + 1)*arctan(a*x)/x, x))*sqrt(c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(a*x)*(c + a^2*c*x^2)^(3/2))/x,x)

[Out]

int((atan(a*x)*(c + a^2*c*x^2)^(3/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}{\left (a x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(3/2)*atan(a*x)/x,x)

[Out]

Integral((c*(a**2*x**2 + 1))**(3/2)*atan(a*x)/x, x)

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